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NEET 2027  ·  Full Syllabus Practice  ·  Aamirah Fathima

Answer Key & Solutions — Set 1

Quick-reference grid first, then a one-line reason for every question with its topic tag. Score the paper, then re-work every wrong answer before moving on.

Questions
180
Max marks
720
Correct
+4
Wrong
−1
Set
1

Quick key

All 180 answers
001 B
002 B
003 B
004 C
005 C
006 C
007 B
008 B
009 C
010 B
011 C
012 B
013 C
014 B
015 C
016 C
017 A
018 C
019 B
020 C
021 C
022 A
023 C
024 A
025 C
026 C
027 B
028 C
029 B
030 B
031 A
032 C
033 C
034 C
035 C
036 B
037 A
038 B
039 B
040 B
041 C
042 B
043 B
044 B
045 C
046 B
047 B
048 D
049 C
050 A
051 D
052 C
053 B
054 B
055 A
056 C
057 B
058 C
059 C
060 A
061 C
062 D
063 C
064 A
065 B
066 A
067 B
068 D
069 C
070 B
071 B
072 C
073 B
074 B
075 C
076 B
077 D
078 A
079 B
080 B
081 B
082 A
083 B
084 D
085 C
086 D
087 B
088 B
089 A
090 D
091 B
092 B
093 B
094 B
095 B
096 D
097 B
098 B
099 C
100 C
101 B
102 B
103 B
104 C
105 B
106 B
107 B
108 C
109 B
110 B
111 B
112 C
113 B
114 C
115 C
116 A
117 B
118 D
119 B
120 C
121 B
122 C
123 C
124 C
125 B
126 B
127 B
128 B
129 B
130 B
131 C
132 B
133 B
134 A
135 B
136 C
137 A
138 B
139 B
140 B
141 B
142 D
143 C
144 B
145 B
146 B
147 C
148 A
149 C
150 C
151 B
152 A
153 B
154 B
155 B
156 B
157 B
158 B
159 B
160 B
161 C
162 C
163 C
164 B
165 B
166 C
167 B
168 C
169 B
170 B
171 B
172 B
173 B
174 B
175 B
176 B
177 B
178 B
179 B
180 B

Physics

Section A  ·  worked reasoning
Q001   (B)   Angular momentum
h has units J·s = kg m² s⁻¹ = [ML²T⁻¹], identical to angular momentum L = mvr.
Units & Dimensions

Q002   (B)   8%
KE = ½mv² ⇒ %error = 2% + 2(3%) = 8%.
Errors

Q003   (B)   45 m
Time up = 3 s, u = 30 m/s, H = u²/2g = 900/20 = 45 m.
Kinematics

Q004   (C)   45°
H/R = tanθ/4 = 1/4 ⇒ tanθ = 1 ⇒ θ = 45°.
Kinematics

Q005   (C)   100 N
F = Δp/Δt = (2×5)/0.1 = 100 N.
Laws of Motion

Q006   (C)   0.58
For equilibrium μ ≥ tanθ = tan30° = 1/√3 ≈ 0.58.
Laws of Motion

Q007   (B)   Comes to rest
In a 1-D elastic collision between equal masses the velocities are exchanged.
Work, Energy & Power

Q008   (B)   10 kW
P = mgh/t = 100×10×10 = 10⁴ W = 10 kW.
Work, Energy & Power

Q009   (C)   4 : 1
(ML²/3)/(ML²/12) = 4 : 1.
Rotational Motion

Q010   (B)   Solid cylinder
a = g sinθ/(1 + k²/R²); solid cylinder has smaller k²/R² (½ vs 1) ⇒ larger a.
Rotational Motion

Q011   (C)   3.0 m
x_cm = (2×0 + 3×5)/5 = 3 m.
Rotational Motion

Q012   (B)   W/2
g_d = g(1 − d/R) = g/2 at d = R/2.
Gravitation

Q013   (C)   8T
T ∝ r^{3/2} ⇒ T' = 4^{3/2} T = 8T.
Gravitation

Q014   (B)   x
Δl = FL/AY; doubling both L and A leaves the ratio L/A unchanged.
Elasticity

Q015   (C)   4T/r
A soap bubble has two surfaces, so ΔP = 4T/r.
Fluids

Q016   (C)   2K₁K₂/(K₁ + K₂)
Thermal resistances add in series ⇒ K_eq = 2K₁K₂/(K₁ + K₂).
Thermal Properties

Q017   (A)   Q = 0 and internal energy decreases
Adiabatic ⇒ Q = 0, so W = −ΔU; gas does work at the cost of internal energy and cools.
Thermodynamics

Q018   (C)   600 K
η = 1 − T₂/T₁ ⇒ 0.5 = 1 − 300/T₁ ⇒ T₁ = 600 K.
Thermodynamics

Q019   (B)   4 : 1
v_rms ∝ 1/√M ⇒ √(32/2) = 4 ⇒ 4 : 1.
Kinetic Theory

Q020   (C)   A/√2
½k(A² − x²) = ½kx² ⇒ x = A/√2.
Oscillations

Q021   (C)   T/2
Each half has 2k; in parallel k_eff = 4k ⇒ T' = T/2.
Oscillations

Q022   (A)   1 : 2
Closed: v/4L, Open: v/2L ⇒ 1 : 2.
Waves

Q023   (C)   1.25 f
f' = f·v/(v − v/5) = 5f/4 = 1.25 f.
Waves

Q024   (A)   Zero
Charge resides on the surface; field inside a conductor is zero.
Electrostatics

Q025   (C)   2pE
W = pE(cos0° − cos180°) = 2pE.
Electrostatics

Q026   (C)   q/6ε₀
Total flux q/ε₀ is shared equally by 6 faces.
Electrostatics

Q027   (B)   V/K
Q constant, C → KC, so V = Q/KC = V/K.
Capacitors

Q028   (C)   One-fourth
v_d = I/(nAe); A ∝ r², so doubling r makes A four times and v_d one-fourth.
Current Electricity

Q029   (B)   1 Ω
r = (E − V)/I = (12 − 10)/2 = 1 Ω.
Current Electricity

Q030   (B)   n²
r ∝ 1/n and B = μ₀nI/2r ⇒ B ∝ n².
Moving Charges & Magnetism

Q031   (A)   2 × 10⁻⁷ N m⁻¹, attractive
F/l = μ₀I₁I₂/2πd = 2 × 10⁻⁷ N/m; parallel currents attract.
Moving Charges & Magnetism

Q032   (C)   Small and negative
Diamagnets have small negative χ, nearly independent of temperature.
Magnetism & Matter

Q033   (C)   1 V
ε = Bvl = 0.5 × 10 × 0.2 = 1 V.
Electromagnetic Induction

Q034   (C)   Four times
L = μ₀n²Al ⇒ L ∝ n².
Electromagnetic Induction

Q035   (C)   Power factor is unity
X_L = X_C ⇒ Z = R (minimum), current maximum, cosφ = 1.
Alternating Current

Q036   (B)   10 turns
Ns = Np·Vs/Vp = 100 × 22/220 = 10.
Alternating Current

Q037   (A)   c
E₀/B₀ = c = 3 × 10⁸ m s⁻¹.
Electromagnetic Waves

Q038   (B)   20 cm
1/f = 1/10 − 1/20 = 1/20 ⇒ f = +20 cm.
Ray Optics

Q039   (B)   42°
sin C = 1/1.5 = 0.667 ⇒ C ≈ 41.8° ≈ 42°.
Ray Optics

Q040   (B)   3β/4
λ decreases by μ ⇒ β' = β/μ = 3β/4.
Wave Optics

Q041   (C)   −3.4 eV
Eₙ = −13.6/n² eV = −13.6/4 = −3.4 eV.
Atoms

Q042   (B)   h/e
eV₀ = hν − φ ⇒ V₀ = (h/e)ν − φ/e, slope = h/e.
Dual Nature of Matter

Q043   (B)   ⁵⁶Fe
The BE/nucleon curve peaks near mass number 56 (≈8.8 MeV for Fe).
Nuclei

Q044   (B)   Decreases
Forward bias opposes the barrier field, so majority carriers narrow the depletion layer.
Semiconductors

Q045   (C)   A NOT gate
Y = (A·A)' = A' — the inverted input, i.e. a NOT gate.
Semiconductors

Chemistry

Section B  ·  worked reasoning
Q046   (B)   6.022 × 10²²
4.4/44 = 0.1 mol ⇒ 0.1 × 6.022 × 10²³ = 6.022 × 10²².
Some Basic Concepts

Q047   (B)   CH₂O
Moles ratio 40/12 : 6.7/1 : 53.3/16 = 3.33 : 6.7 : 3.33 = 1 : 2 : 1.
Some Basic Concepts

Q048   (D)   6
Cr: 3d⁵4s¹ — five 3d and one 4s electron are unpaired.
Structure of Atom

Q049   (C)   10
l = 2 is the d sub-shell: 5 orbitals × 2 electrons = 10.
Structure of Atom

Q050   (A)   Al³⁺ < Mg²⁺ < Na⁺ < F⁻ < O²⁻
For isoelectronic species, radius decreases as nuclear charge increases.
Classification & Periodicity

Q051   (D)   Ne
Ne has a completely filled stable configuration and the highest Z_eff in period 2.
Classification & Periodicity

Q052   (C)   BF₃
BF₃ is trigonal planar and symmetrical, so bond moments cancel.
Chemical Bonding

Q053   (B)   1.5
O₂⁻ has 17 electrons: BO = (10 − 7)/2 = 1.5.
Chemical Bonding

Q054   (B)   O₂
O₂ has two unpaired electrons in its π* antibonding orbitals.
Chemical Bonding

Q055   (A)   7 σ and 3 π
3 C–C sigma + 4 C–H sigma = 7 σ; triple bond gives 2 π and double bond 1 π.
Chemical Bonding

Q056   (C)   ΔH = ΔU − 2RT
Δnɡ = 2 − 4 = −2, so ΔH = ΔU + ΔnɡRT = ΔU − 2RT.
Thermodynamics

Q057   (B)   ΔH < 0, ΔS > 0
ΔG = ΔH − TΔS is negative at every T only when ΔH is negative and ΔS positive.
Thermodynamics

Q058   (C)   Kp = Kc(RT)⁻²
Kp = Kc(RT)^Δn with Δn = −2.
Equilibrium

Q059   (C)   11
pOH = 3 ⇒ pH = 14 − 3 = 11.
Equilibrium

Q060   (A)   pKa
Henderson equation: pH = pKa + log([salt]/[acid]) = pKa when the ratio is 1.
Equilibrium

Q061   (C)   +6
2(+1) + 2x + 7(−2) = 0 ⇒ x = +6.
Redox Reactions

Q062   (D)   C₆H₅CH₂⁺
The benzyl cation is stabilised by resonance with the ring, which outweighs hyperconjugation in t-butyl.
Organic Chemistry – Basic Principles

Q063   (C)   3
Three α-hydrogens on the –CH₃ group give three hyperconjugative structures.
Organic Chemistry – Basic Principles

Q064   (A)   2,4-Dimethylpentane
The longest chain has 5 carbons with methyl groups at C-2 and C-4.
Organic Chemistry – Basic Principles

Q065   (B)   2-Bromopropane
Markovnikov addition places Br on the more substituted carbon via the stabler 2° carbocation.
Hydrocarbons

Q066   (A)   Ethanal only
CH₃CH=CHCH₃ cleaves symmetrically to give two molecules of CH₃CHO.
Hydrocarbons

Q067   (B)   NO₂⁺
H₂SO₄ protonates HNO₃, which loses water to give the nitronium ion NO₂⁺.
Hydrocarbons

Q068   (D)   5
It gives 4 K⁺ and one [Fe(CN)₆]⁴⁻ ⇒ i = 5.
Solutions

Q069   (C)   0.1 m BaCl₂
ΔTb ∝ i × m; BaCl₂ gives i = 3, the largest particle count.
Solutions

Q070   (B)   1.10 V
E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V.
Electrochemistry

Q071   (B)   63.5 g
Cu²⁺ + 2e⁻ → Cu, so 2 F deposits 1 mol = 63.5 g.
Electrochemistry

Q072   (C)   Molar conductivity increases and conductivity decreases
Degree of dissociation rises so Λm increases, while fewer ions per unit volume lower κ.
Electrochemistry

Q073   (B)   100 s
t½ = 0.693/k = 0.693/6.93 × 10⁻³ = 100 s.
Chemical Kinetics

Q074   (B)   Twice the time for 90% completion
t₉₉ = (2.303/k)log100 and t₉₀ = (2.303/k)log10 ⇒ t₉₉ = 2t₉₀.
Chemical Kinetics

Q075   (C)   Sc³⁺
Sc³⁺ is 3d⁰ — no d–d transition is possible.
d- and f-Block Elements

Q076   (B)   Poor shielding by 4f electrons
Diffuse 4f orbitals shield poorly, so effective nuclear charge grows and size shrinks.
d- and f-Block Elements

Q077   (D)   5.92 BM
Fe³⁺ is d⁵ with n = 5 ⇒ μ = √(5×7) = 5.92 BM.
d- and f-Block Elements

Q078   (A)   Pentaamminechloridocobalt(III) chloride
Cobalt is +3; ligands are named alphabetically as ammine before chlorido.
Coordination Compounds

Q079   (B)   d²sp³, 1
CN⁻ is a strong-field ligand: d⁵ pairs to t₂g⁵ ⇒ inner-orbital d²sp³ with one unpaired electron.
Coordination Compounds

Q080   (B)   2
Octahedral MA₄B₂ exists as cis and trans forms only.
Coordination Compounds

Q081   (B)   3° > 2° > 1°
SN1 rate follows carbocation stability, greatest for the tertiary substrate.
Haloalkanes & Haloarenes

Q082   (A)   CH₃Br
SN2 needs backside attack; methyl bromide has the least steric hindrance.
Haloalkanes & Haloarenes

Q083   (B)   Phenol > water > ethanol
Phenoxide is resonance stabilised; ethanol is the weakest acid due to +I of the ethyl group.
Alcohols, Phenols & Ethers

Q084   (D)   2-Methylpropan-2-ol
Tertiary alcohols form the stable 3° carbocation instantly with Lucas reagent.
Alcohols, Phenols & Ethers

Q085   (C)   Benzaldehyde
Benzaldehyde has no α-hydrogen, so no enolate can form.
Aldehydes, Ketones & Carboxylic Acids

Q086   (D)   Pentan-3-one
The test needs a CH₃CO– group (or CH₃CH(OH)–); pentan-3-one has ethyl groups on both sides.
Aldehydes, Ketones & Carboxylic Acids

Q087   (B)   FCH₂COOH > ClCH₂COOH > CH₃COOH
Stronger –I of fluorine stabilises the carboxylate more than chlorine; methyl is electron releasing.
Aldehydes, Ketones & Carboxylic Acids

Q088   (B)   (C₂H₅)₂NH > C₂H₅NH₂ > NH₃ > C₆H₅NH₂
Aniline is weakest (lone pair delocalised); in water the 2° amine is the strongest of the ethylamines.
Amines

Q089   (A)   Primary amines
Only 1° amines with CHCl₃/alcoholic KOH give foul-smelling isocyanides.
Amines

Q090   (D)   Sucrose
In sucrose both anomeric carbons are involved in the glycosidic linkage, so no free aldehyde group remains.
Biomolecules

Botany

Section C  ·  worked reasoning
Q091   (B)   Specific epithet
Binomial nomenclature: first word is the genus, second is the specific epithet.
The Living World

Q092   (B)   Mycoplasma
Mycoplasmas are the smallest living cells and can survive without oxygen; they have no cell wall.
Biological Classification

Q093   (B)   Dinoflagellates
Red dinoflagellates such as Gonyaulax multiply rapidly and make the sea appear red.
Biological Classification

Q094   (B)   Ascomycetes and Basidiomycetes
In both Ascomycetes and Basidiomycetes plasmogamy is followed by a prolonged dikaryophase.
Biological Classification

Q095   (B)   A haploid gametophyte
The fern prothallus is the free-living, photosynthetic haploid gametophyte.
Plant Kingdom

Q096   (D)   Angiosperms
Only angiosperms show syngamy plus triple fusion within the same embryo sac.
Plant Kingdom

Q097   (B)   Twisted
In twisted aestivation one margin of each petal overlaps the next in a regular direction.
Morphology of Flowering Plants

Q098   (B)   Dianthus
In Dianthus and Primrose ovules are borne on a central axis with no septa.
Morphology of Flowering Plants

Q099   (C)   Fabaceae
Fabaceae: papilionaceous corolla, vexillary aestivation, stamens (9) + 1, marginal placentation.
Morphology of Flowering Plants

Q100   (C)   Endodermis
Suberin deposits on radial and tangential walls of endodermal cells form Casparian strips.
Anatomy of Flowering Plants

Q101   (B)   Conjoint, closed and scattered
Monocot stems have scattered, closed (no cambium) conjoint bundles.
Anatomy of Flowering Plants

Q102   (B)   Adaxial epidermis of grass leaves
These large empty cells in grass leaf upper epidermis help the leaf curl inwards to reduce water loss.
Anatomy of Flowering Plants

Q103   (B)   70S with 50S and 30S subunits
Bacterial ribosomes are 70S, built from 50S and 30S subunits.
Cell: The Unit of Life

Q104   (C)   Calcium pectate
Calcium pectate cements neighbouring cells together.
Cell: The Unit of Life

Q105   (B)   Ribosome
Ribosomes are naked ribonucleoprotein particles with no surrounding membrane.
Cell: The Unit of Life

Q106   (B)   Competitive inhibitor
Malonate resembles succinate in structure and competes for the active site.
Biomolecules

Q107   (B)   Rubber
Rubber, alkaloids, essential oils and gums are secondary metabolites of plants.
Biomolecules

Q108   (C)   Pachytene
Recombination nodules appear at pachytene and recombinase mediates crossing over.
Cell Cycle & Cell Division

Q109   (B)   Zygotene
Synapsis of homologous chromosomes at zygotene forms the synaptonemal complex.
Cell Cycle & Cell Division

Q110   (B)   Oxaloacetic acid
PEP carboxylase in mesophyll cells fixes CO₂ into the 4-carbon OAA.
Photosynthesis

Q111   (B)   18 ATP and 12 NADPH
Each CO₂ needs 3 ATP and 2 NADPH ⇒ 18 ATP and 12 NADPH for six.
Photosynthesis

Q112   (C)   There is no synthesis of sugar or ATP
The C₂ pathway releases CO₂ without producing ATP, NADPH or carbohydrate.
Photosynthesis

Q113   (B)   P700
PS I absorbs maximally at 700 nm and its reaction centre is P700.
Photosynthesis

Q114   (C)   36
NCERT's balance sheet gives a net of 36 ATP per glucose under ideal assumptions.
Respiration in Plants

Q115   (C)   Less than 1
Fats are poor in oxygen and consume more O₂, giving RQ around 0.7.
Respiration in Plants

Q116   (A)   Auxin
Auxins promote fruit development without fertilisation.
Plant Growth & Development

Q117   (B)   Vernalisation
Vernalisation prevents premature flowering and depends on a cold treatment.
Plant Growth & Development

Q118   (D)   Tapetum
Tapetal cells are dense, multinucleate and supply nutrition to microspores.
Sexual Reproduction in Flowering Plants

Q119   (B)   8-nucleate and 7-celled
Three successive free nuclear divisions give 8 nuclei organised into 7 cells.
Sexual Reproduction in Flowering Plants

Q120   (C)   Triploid
Triple fusion of one male gamete with two polar nuclei gives a triploid PEN.
Sexual Reproduction in Flowering Plants

Q121   (B)   Nucellar cells developing into embryos
Nucellar cells protrude into the embryo sac and form adventive embryos.
Sexual Reproduction in Flowering Plants

Q122   (C)   1 : 1 : 1 : 1
Crossing F₁ with the double recessive parent gives four equally frequent classes.
Principles of Inheritance

Q123   (C)   Multiple allelism and codominance
Three alleles I^A, I^B and i exist, and I^A and I^B are codominant in AB individuals.
Principles of Inheritance

Q124   (C)   50%
Sons receive the X from the mother; half of her X chromosomes carry the recessive allele.
Principles of Inheritance

Q125   (B)   Trisomy of chromosome 21
An additional copy of chromosome 21 gives a total of 47 chromosomes.
Principles of Inheritance

Q126   (B)   Avery, MacLeod and McCarty
They showed that only DNase abolished transformation, proving DNA is the transforming principle.
Molecular Basis of Inheritance

Q127   (B)   DNA ligase
Ligase seals the nicks between the discontinuously synthesised fragments.
Molecular Basis of Inheritance

Q128   (B)   Allolactose
Allolactose binds the repressor produced by the i gene and inactivates it.
Molecular Basis of Inheritance

Q129   (B)   3.3 × 10⁹
The Human Genome Project reported about 3164.7 million base pairs.
Molecular Basis of Inheritance

Q130   (B)   Sigmoid
Resource limitation gives the Verhulst–Pearl S-shaped curve with a carrying capacity K.
Organisms & Populations

Q131   (C)   Mutualism
The alga supplies food and the fungus supplies shelter, water and minerals — both benefit.
Organisms & Populations

Q132   (B)   Always upright
Energy is lost at every transfer, so the energy pyramid can never be inverted.
Ecosystem

Q133   (B)   GPP − respiration losses
NPP = GPP − R, and it is the biomass available to herbivores and decomposers.
Ecosystem

Q134   (A)   0.1 and 0.2
The regression slope Z is remarkably similar, 0.1–0.2, regardless of taxonomic group.
Biodiversity & Conservation

Q135   (B)   In situ conservation
Sacred groves protect species in their natural habitat, hence in situ conservation.
Biodiversity & Conservation

Zoology

Section D  ·  worked reasoning
Q136   (C)   Echinodermata
Echinoderms use the water vascular system with tube feet for locomotion and feeding.
Animal Kingdom

Q137   (A)   Platyhelminthes
Protonephridia or flame cells carry out osmoregulation and excretion in flatworms.
Animal Kingdom

Q138   (B)   Arthropoda
In arthropods, blood flows through a haemocoel rather than closed vessels.
Animal Kingdom

Q139   (B)   Cephalochordata
In Cephalochordata (Branchiostoma) the notochord extends from head to tail throughout life.
Animal Kingdom

Q140   (B)   Loose connective tissue
Areolar tissue is loose connective tissue that acts as a support framework beneath the skin.
Structural Organisation in Animals

Q141   (B)   Malpighian tubules
About 100–150 yellow Malpighian tubules at the junction of midgut and hindgut remove nitrogenous waste.
Structural Organisation in Animals

Q142   (D)   pCO₂ and H⁺ increase
Bohr effect: high CO₂, high H⁺ and high temperature favour dissociation of oxyhaemoglobin.
Breathing & Exchange of Gases

Q143   (C)   70%
About 70% of CO₂ is carried as HCO₃⁻, 20–25% as carbamino-haemoglobin and 7% dissolved.
Breathing & Exchange of Gases

Q144   (B)   TV + IRV + ERV
Vital capacity is the maximum volume that can be exhaled after a maximal inspiration.
Breathing & Exchange of Gases

Q145   (B)   SA node
The SA node in the right atrium generates about 70–75 impulses per minute.
Body Fluids & Circulation

Q146   (B)   Fibrinogen to fibrin
Thrombin catalyses conversion of soluble fibrinogen into insoluble fibrin threads.
Body Fluids & Circulation

Q147   (C)   Atrioventricular valves
Closure of the bicuspid and tricuspid valves at the start of ventricular systole gives 'lubb'.
Body Fluids & Circulation

Q148   (A)   Proximal convoluted tubule
Nearly all essential nutrients and about 70–80% of electrolytes and water are reclaimed in the PCT.
Excretory Products

Q149   (C)   Distal tubule and collecting duct
ADH increases facultative water reabsorption in the DCT and collecting duct.
Excretory Products

Q150   (C)   Uricotelic
They excrete uric acid as a semi-solid paste, conserving water.
Excretory Products

Q151   (B)   The I band shortens
Actin filaments slide inwards, so the I band narrows while the A band length is unchanged.
Locomotion & Movement

Q152   (A)   80
Skull 22 + hyoid 1 + ear ossicles 6 + vertebral column 26 + sternum 1 + ribs 24 = 80.
Locomotion & Movement

Q153   (B)   Uric acid crystals
Inflammation of joints follows deposition of uric acid crystals.
Locomotion & Movement

Q154   (B)   The Na⁺/K⁺ pump moving 3 Na⁺ out and 2 K⁺ in
The electrogenic sodium–potassium pump keeps the inside negative at about −70 mV.
Neural Control & Coordination

Q155   (B)   From one node of Ranvier to the next
In myelinated fibres, depolarisation jumps between nodes of Ranvier, speeding conduction.
Neural Control & Coordination

Q156   (B)   Cerebellum
The cerebellum coordinates precise voluntary movement and body equilibrium.
Neural Control & Coordination

Q157   (B)   Beta cells of islets of Langerhans
Beta cells secrete insulin; alpha cells secrete glucagon.
Chemical Coordination

Q158   (B)   Synthesised by the hypothalamus and released from the posterior pituitary
The neurohypophysis stores and releases hormones made by hypothalamic neurons.
Chemical Coordination

Q159   (B)   Hyposecretion of the adrenal cortex
Deficiency of glucocorticoids and mineralocorticoids causes Addison's disease.
Chemical Coordination

Q160   (B)   Provide nutrition to germ cells
Sertoli cells nourish germ cells; Leydig cells secrete androgens.
Human Reproduction

Q161   (C)   Ampullary–isthmic junction of the fallopian tube
Sperm and ovum meet at the ampullary–isthmic junction of the oviduct.
Human Reproduction

Q162   (C)   Human chorionic gonadotropin
hCG from the trophoblast sustains the corpus luteum, which secretes progesterone.
Human Reproduction

Q163   (C)   1–2 million
About one to two million primary oocytes per ovary are present at birth.
Human Reproduction

Q164   (B)   Suppressing sperm motility and fertilising capacity
Cu ions released by Cu-T and Multiload suppress sperm motility and fertilising capacity.
Reproductive Health

Q165   (B)   12 weeks
MTPs are comparatively safe during the first trimester, i.e. up to 12 weeks.
Reproductive Health

Q166   (C)   A zygote or early embryo up to the 8-blastomere stage
Zygote intrafallopian transfer places the zygote or early embryo into the fallopian tube.
Reproductive Health

Q167   (B)   Natural selection
Soot-darkened tree trunks favoured the dark moths, which then increased in frequency.
Evolution

Q168   (C)   0.48
q = 0.4, p = 0.6 ⇒ 2pq = 2 × 0.6 × 0.4 = 0.48.
Evolution

Q169   (B)   Analogous and show convergent evolution
Different structural origins but the same function — analogy from convergent evolution.
Evolution

Q170   (B)   650–800 cc
Homo habilis had a brain capacity of about 650–800 cc and probably did not eat meat.
Evolution

Q171   (B)   Haemozoin
Haemozoin, a toxin released when infected RBCs rupture, produces the periodic chill and fever.
Human Health & Disease

Q172   (B)   Dengue
Dengue is viral; typhoid is bacterial, malaria protozoan and ringworm fungal.
Human Health & Disease

Q173   (B)   H₂L₂
Each antibody has two heavy and two light polypeptide chains.
Human Health & Disease

Q174   (B)   Helper T lymphocytes
After replicating in macrophages, HIV progressively destroys helper T cells.
Human Health & Disease

Q175   (B)   GAATTC
EcoRI cuts between G and A in the palindrome GAATTC, producing sticky ends.
Biotechnology – Principles

Q176   (B)   Taq polymerase
Taq polymerase from Thermus aquaticus remains active during denaturation at ~94 °C.
Biotechnology – Principles

Q177   (B)   Controls the copy number of the linked DNA
The origin of replication initiates replication and governs the copy number.
Biotechnology – Principles

Q178   (B)   In the alkaline gut of the insect
The alkaline pH of the insect gut solubilises the crystal and activates the toxin.
Biotechnology – Applications

Q179   (B)   Meloidogyne incognita
dsRNA silences a specific nematode mRNA, protecting tobacco from Meloidogyne incognita.
Biotechnology – Applications

Q180   (B)   Adenosine deaminase deficiency
ADA deficiency was treated by infusing lymphocytes carrying a functional ADA cDNA.
Biotechnology – Applications